YES

We show the termination of the TRS R:

  div(X,e()) -> i(X)
  i(div(X,Y)) -> div(Y,X)
  div(div(X,Y),Z) -> div(Y,div(i(X),Z))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(X,e()) -> i#(X)
p2: i#(div(X,Y)) -> div#(Y,X)
p3: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))
p4: div#(div(X,Y),Z) -> div#(i(X),Z)
p5: div#(div(X,Y),Z) -> i#(X)

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(X,e()) -> i#(X)
p2: i#(div(X,Y)) -> div#(Y,X)
p3: div#(div(X,Y),Z) -> i#(X)
p4: div#(div(X,Y),Z) -> div#(i(X),Z)
p5: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (3,2)
            e_A() = (2,2)
            i#_A(x1) = ((0,0),(1,0)) x1 + (3,3)
            div_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,5)
            i_A(x1) = ((1,0),(1,1)) x1 + (0,1)
    
      precedence:
      
        i# > div# = e = div = i
      
      partial status:
    
        pi(div#) = []
        pi(e) = []
        pi(i#) = []
        pi(div) = [1]
        pi(i) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = (2,1)
            e_A() = (1,0)
            i#_A(x1) = (0,0)
            div_A(x1,x2) = (2,0)
            i_A(x1) = (2,0)
    
      precedence:
      
        div# = div = i > e = i#
      
      partial status:
    
        pi(div#) = []
        pi(e) = []
        pi(i#) = []
        pi(div) = []
        pi(i) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(X,e()) -> i#(X)
p2: div#(div(X,Y),Z) -> i#(X)
p3: div#(div(X,Y),Z) -> div#(i(X),Z)
p4: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The estimated dependency graph contains the following SCCs:

  {p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(div(X,Y),Z) -> div#(i(X),Z)
p2: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,2)
            div_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
            i_A(x1) = ((1,0),(1,1)) x1
            e_A() = (1,0)
    
      precedence:
      
        i > div# = div > e
      
      partial status:
    
        pi(div#) = [1]
        pi(div) = [2]
        pi(i) = []
        pi(e) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = x1 + (1,1)
            div_A(x1,x2) = x2 + (2,2)
            i_A(x1) = (0,0)
            e_A() = (0,1)
    
      precedence:
      
        i > div > div# > e
      
      partial status:
    
        pi(div#) = [1]
        pi(div) = []
        pi(i) = []
        pi(e) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: div#(div(X,Y),Z) -> div#(Y,div(i(X),Z))

and R consists of:

r1: div(X,e()) -> i(X)
r2: i(div(X,Y)) -> div(Y,X)
r3: div(div(X,Y),Z) -> div(Y,div(i(X),Z))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
            div_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (3,3)
            i_A(x1) = ((1,0),(1,1)) x1 + (0,2)
            e_A() = (1,1)
    
      precedence:
      
        div# = div = i = e
      
      partial status:
    
        pi(div#) = []
        pi(div) = []
        pi(i) = []
        pi(e) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            div#_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
            div_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,1)
            i_A(x1) = x1 + (2,3)
            e_A() = (1,2)
    
      precedence:
      
        div# = div = i = e
      
      partial status:
    
        pi(div#) = []
        pi(div) = []
        pi(i) = [1]
        pi(e) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.